- #1
timsea81
- 89
- 1
du/dt = A*d^2u/dx^2
Where u=u(x,t) and A is a constant.
The boundary conditions are:
u=f(x) when t=0
u=0 when x=0, independent of t
u=V when x=L, independent of t, where V and L are both non-zero constants.
Is there a general solution to this, or do I need to solve it numerically?
I tried solving it by separation of variables similar to what was done here:
http://en.wikipedia.org/wiki/Heat_equation#Solving_the_heat_equation_using_Fourier_series
But in the link the last boundary condition is u=0 when x=L, which allows us to conclude something about lambda that I can't with a non-zero boundary condition.
If it is separable, we have u(x,t)=X(x)T(t)=V when x=L independent of t. Following the method used in the link, to the point where we apply the different boundary conditions and my problem becomes different than what is in the link:
u(x,t) = [B sin(rootlambda x)] * [A e^-(lambda alpha t)]
Since u(L,t) is a non-zero constant independent of t, I think that I need to conclude A=0 leading to a trivial solution. I got this far with it before and concluded that the solution was not separable and therefore can only be evaluated numerically. Do you agree, or am I missing something?
Here is a link to another forum where I am looking for the same answer:
http://www.reddit.com/r/cheatatmathhomework/comments/17i341/dudt_ad2udy2/
Where u=u(x,t) and A is a constant.
The boundary conditions are:
u=f(x) when t=0
u=0 when x=0, independent of t
u=V when x=L, independent of t, where V and L are both non-zero constants.
Is there a general solution to this, or do I need to solve it numerically?
I tried solving it by separation of variables similar to what was done here:
http://en.wikipedia.org/wiki/Heat_equation#Solving_the_heat_equation_using_Fourier_series
But in the link the last boundary condition is u=0 when x=L, which allows us to conclude something about lambda that I can't with a non-zero boundary condition.
If it is separable, we have u(x,t)=X(x)T(t)=V when x=L independent of t. Following the method used in the link, to the point where we apply the different boundary conditions and my problem becomes different than what is in the link:
u(x,t) = [B sin(rootlambda x)] * [A e^-(lambda alpha t)]
Since u(L,t) is a non-zero constant independent of t, I think that I need to conclude A=0 leading to a trivial solution. I got this far with it before and concluded that the solution was not separable and therefore can only be evaluated numerically. Do you agree, or am I missing something?
Here is a link to another forum where I am looking for the same answer:
http://www.reddit.com/r/cheatatmathhomework/comments/17i341/dudt_ad2udy2/